Heuristics for the two-stage job shop scheduling problem with a bottleneck machine

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1 European Journal of Operational Research 123 (2000) 229±240 Heuristics for the two-stage job shop scheduling problem with a bottleneck machine I.G. Drobouchevitch, V.A. Strusevich * School of Computing and Mathematical Sciences, University of Greenwich, Maritime Greenwich University Campus, 30 Park Row, London SE10 9LS, UK Abstract The paper considers the job shop scheduling problem to minimize the makespan. It is assumed that each job consists of at most two operations, one of which is to be processed on one of m P 2 machines, while the other operation must be performed on a single bottleneck machine, the same for all jobs. For this strongly NP-hard problem we present two heuristics with improved worst-case performance. One of them guarantees a worst-case performance ratio of 3=2. The other algorithm creates a schedule with the makespan that exceeds the largest machine workload by at most the length of the largest operation. Ó 2000 Elsevier Science B.V. All rights reserved. Keywords: Job shop scheduling; Approximation algorithm; Worst-case analysis 1. Introduction * Corresponding author. Tel.: ; fax: address: v.strusevich@greenwich.ac.uk (V.A. Strusevich). In the general job shop scheduling model, each job consists of several operations to be processed on all or some of the given machines. The operations of a job have to follow the assigned processing route, speci c for each job. In such a route, some of the machines may be skipped, while the others may appear more than once. If all jobs have the same processing route, the model is called the ow shop. Given the processing times of all operations, a purpose is to nd a schedule that minimizes a certain objective function. In all problems considered in this paper the objective function to be minimized is the makespan, i.e., the maximum completion time of all jobs on all machines. It is often found in practice that the processing routes may be short, e.g., consist of two or three operations each. (See [6,8,14] for a discussion of various applications of the two-stage shop scheduling problems.) In this paper we consider the job shop scheduling problem in which the processing route for each job consists of two operations at most. Moreover, in our model the number of machines is arbitrary, and one of the operations of each job has to be processed on a particular machine, the same for all jobs. We call such a machine the /00/$ - see front matter Ó 2000 Elsevier Science B.V. All rights reserved. PII: S ( 9 9 )

2 230 I.G. Drobouchevitch, V.A. Strusevich / European Journal of Operational Research 123 (2000) 229±240 bottleneck machine. In many applications the bottleneck machine can be seen as an input/output bay, where the orders arrive to be further ful lled or where the completed orders are gathered before shipped to the customer. The model under consideration is closely related to some other scheduling problems known in the literature, such as the two-stage ow shop problem with parallel machines in one stage [1,12] and the two-stage assembly scheduling problem [15]. For these and similar shop scheduling problems, research has recently been focused on developing and analyzing approximation algorithms. Because of their relative simplicity, these models have appeared to be helpful for testing various scheduling techniques. On the other hand, even the short route problems are not straightforward to solve. It is known that for the job shop with three operations per job nding a schedule close to an optimal one is no easier than solving the problem to optimality [19]. In this paper we extend the approximation results obtained in [6,8,13,14], where the ow shop counterpart of our problem is studied. Our algorithms not only handle a more general job shop problem, but also guarantee a better performance than heuristics known for the ow shop. The remainder of the paper is organized as follows. Section 2 gives a formal description of the problem. Section 3 brie y overviews the relevant results in the area. In Section 4 we study properties of the job shop schedules with the bottleneck machine. In Section 5 we present an algorithm that creates a schedule with the makespan that is no more than 3/2 times the optimal value. Section 6 shows how to nd a schedule with the makespan that exceeds the largest machine workload by at most the length of the largest operation. 2. Problem description We start with a formal description of the job shop. We are given m machines and a set of n jobs, where job j consists of m j operations. Each operation has to be processed on a speci ed machine. The processing times of all operations are known. For each job, the order in which its operations must be processed is known in advance, and can be di erent for di erent jobs. There is no preemption in processing an operation. The usual scheduling requirement says that no two jobs may be processed on a machine at a time, and each job may not be assigned to more than one machine at a time. For a schedule S the value of the makespan is denoted by C max S. To refer to a particular scheduling problem we follow a standard three- eld notation as in [11]. If the number of machines is variable, the job shop problem is denoted by JjjC max. If the number of machines m is xed and the length of any route does not exceed a given value of m 0, then the problem is denoted by Jmjm j 6 m 0 jc max. The J2jm j 6 2jC max problem is solvable in O n log n time [9]. If for each job the processing route is the same and no machine is visited twice, we obtain the ow shop problem traditionally denoted by F jjc max. The well-known Johnson algorithm solves the F 2jjC max problem in O n log n time. The F 3jjC max problem is strongly NP-hard as proved in [4]. We now formally describe the two-stage job shop scheduling problem with a single bottleneck machine, which is the main subject of this study. We are given m 1 machines, m P 2; denoted by A 1 ; A 2 ;...; A m and B. Each job j consists of at most two operations, one of which is processed on the bottleneck machine B: For each machine A k,we are given a set N k that consists of all jobs that are rst processed on A k and then on B; in other words, all jobs of N k are given the processing route A k ; B. Similarly, for A k we are given a set H k of all jobs that must follow the opposite route B; A k.in any case, for a job j 2 N k [ H k its processing on machine A k takes a j time units, while the processing time on the bottleneck machine B is equal to b j. The jobs to be processed on machine B can be grouped into two sets N ˆ Sm kˆ1 N k and H ˆ Sm kˆ1 H k with jnj ˆn and jhj ˆh: We will denote the problem of minimizing the makespan for the described processing system by J m 1 jm j 6 2jC max. If all jobs have the same route, e.g., if H ˆ;, the resulting ow shop problem is denoted by F m 1 jm j 6 2jC max.as shown in [8,13], the F m 1 jm j 6 2jC max problem is NP-hard in the strong sense if m ˆ 2. This

3 I.G. Drobouchevitch, V.A. Strusevich / European Journal of Operational Research 123 (2000) 229± implies the same complexity status of the corresponding job shop problem J m 1 jm j 6 2jC max for m P 2. In the F m 1 jm j 6 2jC max problem a job j has to be processed rst on a particular machine A k : If, otherwise, each job j may be processed on any of the machines A 1 ; A 2 ;...; A m and this still takes a j time, we obtain the two-stage ow shop problem with m identical parallel machines in the rst stage [1]. On the other hand, the F m 1 jm j 6 2jC max problem is a special case of the two-stage assembly problem, provided that B is the assembly machine and each job has to be processed on exactly one machine in the rst stage [15]. Since the J m 1 jm j 6 2jC max problem and most of the relevant problems are NP-hard, nding an optimal solution in reasonable computation time is highly unlikely. This initiates the search for fast approximation algorithms that may nd a schedule close to an optimal one. Section 3 gives a brief overview of the results in this area. 3. Review of shop approximation For a schedule created by an approximation algorithm it is desirable to be able to know in advance how close to the optimum that schedule appears to be. The quality of S is usually measured by the ratio C max S =C max S or by the di erence C max S C max S, where S is an optimal schedule for the given problem instance. An algorithm H that creates a schedule S H is called a q-approximation algorithm if the inequality C max S H =C max S 6 q holds for all problem instances. The smallest possible value of q is called the worst-case performance ratio of the algorithm. As shown by Hall [7], the FmjjC max problem admits a polynomial-time approximation scheme (PTAS), i.e., there exists a q-approximation algorithm with q ˆ 1 e for any given e > 0, and the running time of such an algorithm is polynomial for xed m and e. This approach cannot be extended for the general JmjjC max problem. However, a similar argument carries over for the J m 1 jm j 6 2jC max problem because of its speci c features. Therefore, it can be concluded that the J m 1 jm j 6 2jC max problem admits a PTAS, though its running time appears highly impractical. On the other hand, even the F m 1 jm j 6 2jC max problem with m ˆ 2 does not allow a fully polynomial approximation scheme (unless P ˆ NP), since the problem is strongly NP-hard. Recall that an approximation scheme is called fully polynomial if q ˆ 1 e for any given e > 0, and the running time is polynomial in both the length of the problem input and 1=e. For the JmjjC max problem a schedule S e such that C max S e C max S 6 2 e holds for any given e > 0, can be found in time that is polynomial for xed m and 1=e [18]. This result and the PTAS mentioned above produce important theoretical evidence of the existence of polynomial algorithms with small values of q. On the other hand, Williamson et al. [19] show that, unless P ˆ NP, there is no polynomial time algorithm for the F jjc max problem which provides q < 5=4. Notice that the latter result only holds if the instance of the problem contains a job with at least three operations. For the FmjjC max and Jmjm j 6 m 0 jc max problems, heuristic algorithms that require reasonable computational e ort are not known to provide even a constant worst-case performance ratio q [5]. For F 3jjC max problem, a 5=3-approximation algorithm is developed in [2]. For each of the J3jm j 6 2jC max and J2jm j 6 3jC max problems, 3=2-approximation algorithms are o ered in [3]. For the two-stage assembly problem with m machines in the rst stage, Potts et al. [15] give a number of heuristics the best of which yields q ˆ 2 1=m: The same ratio is known for the twostage ow shop problem with at most m parallel machines in a stage [12]. For the F m 1 jm j 6 2jC max problem with m ˆ 2, algorithms with q ˆ 3=2 are described in [8,14]. That result is extended by Gupta [6], who designs several heuristic algorithms that provide q ˆ 2 1=m for the F m 1 jm j 6 2jC max problem. In this paper, we give a 3=2-approximation algorithm for the general J m 1 jm j 6 2jC max problem.

4 232 I.G. Drobouchevitch, V.A. Strusevich / European Journal of Operational Research 123 (2000) 229±240 For a shop scheduling problem, let P denote the largest machine workload, i.e., the sum of the processing times of all operations assigned to the machine. The length of the largest operation is denoted by p max. For various shop scheduling problems, there are heuristic algorithms that create schedule S H such that C max S H 6 P ap max. Observe that the value ap max can be considered as a bound on the absolute error C max S H C max S, since C max S P P. This also implies that C max S 2 P ; P ap max Š: 1 Given a scheduling problem, it is an interesting research goal to nd the smallest value of a for which (1) holds. Most of the results in this direction belong to Sevast'janov and usually employ geometric methods, such as compact vector summation. See [16,17] for the recent developments in this area. In this paper, we show how to nd a schedule S for the J m 1 jm j 6 2jC max problem such that C max S 6 P p max. This extends an analogous result established for the F m 1 jm j 6 2jC max problem with m ˆ 2 [13]. 4. Finding job shop schedules In this section we present algorithms for the ow shop and job shop scheduling problems with a single bottleneck machine. The algorithms provide the basis for our heuristics introduced in the subsequent sections. The algorithms employ the idea of greedy scheduling on the bottleneck machine. Given the J m 1 jm j 6 2jC max problem, we distinguish between the operations of a job, referring to them as the rst-stage operation and the second-stage operation, respectively. Thus, for a job of some set N k ; its rst-stage operation is performed on machine A k ; while its second-stage operation is done on the bottleneck machine. For the jobs of a set H k the operations are classi ed analogously. For our purposes, we may restrict our attention to the schedules with the following structural properties: (i) Every machine A k ; 1 6 k 6 m; rst processes all jobs of set N k followed by all jobs of set H k : (ii) The bottleneck machine starts with all jobs of set H followed by all jobs of set N. (iii) On each machine the rst-stage operations may start at time zero and be processed as a block, with no intermediate idle time. (iv) For the jobs of each set N k or H k ; the rststage operations and the second-stage operations follow the same sequence. Due to standard interchange argument, any schedule can be converted into the one of the required structure without increasing the makespan. For an non-empty set Q N [ H we de ne a Q ˆX a j ; b Q ˆX b j ; j2q j2q assuming that a ; ˆ b ; ˆ 0: For a schedule S, the starting time and the completion time of a job j on a machine A k are denoted by R j;a S and C j;a S, respectively. For the bottleneck machine, similar values are denoted by R j;b S and C j;b S, respectively. If no confusion arises, we may drop the reference to the schedule and write R j;a or C j;b. We now concentrate on the ow shop F m 1 jm j 6 2jC max problem of scheduling the jobs of set N only. Let u k N k denote a given sequence of jobs of set N k on machine A k. Our current purpose is to describe an algorithm that takes these sequences as an input and nds a feasible schedule satisfying the above properties. What needs to be determined is the sequence u N of all operations and their starting times (or completion times) on the bottleneck machine. Consider a feasible ow shop schedule S 0 that is terminated on machine B. Assume that there is no unforced idle time on B, i.e., machine B is idle if and only if there is no job that may be started on it. There exists a job w such that: (a) the second-stage operation of w starts exactly at time its rst-stage operation completes, i.e., C w;a S 0 ˆR w;b S 0 ; (b) job w is the rst job in the sequence u N after which machine B has no idle time. Job w is called critical. We cannot increase the starting time R w;b S 0 of the critical job without increasing the makespan, as long as the sequence

5 I.G. Drobouchevitch, V.A. Strusevich / European Journal of Operational Research 123 (2000) 229± u N is kept. For our purposes, it is convenient to transform S 0 into another schedule S with R w;b S ˆR w;b S 0 and with the same makespan by letting all jobs that precede job w in u N start on B as late as possible. In the resulting schedule S, the sequence u N of jobs starts on B at time R B S ˆC max S b N and the jobs are processed as a block, without intermediate idle time. The algorithm presented below nds a ow shop schedule S, which is the best of all feasible schedules associated with a given collection of the sequences of the rst-stage operations. Algorithm FSN. Input: An instance of the F m 1 jm j 6 2jC max problem and the sequences u 1 N 1, u 2 N 2 ;...; u m N m. Output: Feasible schedule S. 1. For each machine A k start the sequence u k N k at time zero and process the jobs of N k as a block, with no intermediate idle time. 2. Scanning the jobs in the order they are completed in the rst stage, nd schedule S 0 by starting the processing of the next available job on machine B as early as possible, until all jobs are scheduled. Call the resulting sequence u N. Compute the value of the makespan C max S 0 as the completion time of the last job in u N on machine B. 3. Compute Y B S ˆC max S 0 b N. In the found schedule S, the sequence u N of jobs starts on B at time Y B S ; machine B has no intermediate idle time and C max S ˆC max S 0 : Let w be the critical job in schedule S 0. It is clear that w remains critical in S, and R w;b S ˆR w;b S 0. Algorithm FSN assigns the jobs to machine B in the greedy manner. In any feasible schedule associated with a given collection of the permutations u 1 N 1, u 2 N 2 ;...; u m N m, neither job w nor any of the jobs that follow w in the sequence u N may start on B earlier than time R w;b S. Therefore, the found schedule S provides the smallest makespan among all ow shop schedules related to the permutations u 1 N 1, u 2 N 2 ;...; u m N m : We now pass to the ow shop problem of processing the jobs of set H, that is symmetric to the one considered above. For each k, 16 k 6 m; assume that the jobs of set H k have to be processed on machine A k according to a given sequence w k H k. Algorithm FSN can be modi ed to nd the best schedule among those associated with the given permutations. Algorithm FSH. Input: An instance of the ow shop problem for processing the jobs of set H and a collection of the sequences w 1 H 1, w 2 H 2 ;...; w m H m. Output: Feasible schedule S. 1. For each k; 1 6 k 6 m; nd the permutation ew k H k obtained by inverting the order of jobs in the permutation w k H k. Introduce the arti cial F m 1 jm j 6 2jC max problem with the set of jobs H, assuming that for all these jobs the second-stage operation is processed on the bottleneck machine B. 2. Run Algorithm FSN for the obtained F m 1 jm j 6 2jC max problem with w e k H k ; k ˆ 1; 2;...; m. Let es be the found schedule in which machine B processes the jobs of H in the sequence w H : e 3. Find the permutation w H by taking the jobs in w H e in the opposite order. Transform es into a feasible schedule S for the original problem by inverting time, so that C max es ˆC max S. In the found schedule S, the jobs of set H start on B at time zero and follow the sequence w H, while each of the machines A k starts at time Yk A S ˆC max es a H k and processes the jobs of set H k in the sequence w k H k with no intermediate idle time. We now turn to the original J m 1 jm j 6 2jC max problem. Notice that the problem can be viewed as two independent ow shop problems with the same bottleneck machine: one for processing the jobs of set N, and the other for processing the jobs in H. For each k, 16 k 6 m; assume that for machine A k the jobs of set N k must follow a given sequence u k N k, while the jobs of set H k have to be processed according to a given sequence w k H k. We describe how the ow shop schedules found by Algorithm FSN and Algorithm FSH can be combined together to produce the desired job shop schedule. If the direct superposition of the two schedules produces clashes, i.e., on some machine the rst-stage operations

6 234 I.G. Drobouchevitch, V.A. Strusevich / European Journal of Operational Research 123 (2000) 229±240 complete later that the second-stage operations start, then we nd the longest interval of such an overlap and shift the con icting second-stage operations to the right. Algorithm JS. Input: An instance of the J m 1 jm j 6 2jC max problem and two collections of the sequences u 1 N 1, u 2 N 2 ;...; u m N m and w 1 H 1, w 2 H 2 ;...; w m H m. Output: Feasible schedule S. 1. Run Algorithm FSN for the ow shop problem with the set of jobs N. Let S N be the found schedule in which machine B starts processing the jobs in set N at time Y B S N : 2. Run Algorithm FSH for the ow shop problem with the set of jobs H. For the found schedule S H, a machine A k starts processing the jobs in set H k at time Yk A S H ; k ˆ 1; 2;...; m. 3. For each k; k ˆ 1; 2;...; m; compute D k ˆ max a N k Yk A S H ; 0. Compute DB ˆ max fb H Y B S N ; 0g: 4. In schedule S N ; increase the starting time of each operation on machine B by D B. In schedule S H ; increase the starting time of each operation on every machine A k by D k. Combine the resulting schedules together to obtain the required schedule S: Notice that in schedule created by Algorithm JS a machine for which the shifts in step 4 have been made starts at time zero and has no intermediate idle time. The value of the makespan of the resulting schedule either is equal to the workload of some machine or is determined by the makespan of one of the schedules S N or S H. The running time of each of the Algorithms FSN, FSH and JS is linear in the number of jobs. 5. Heuristic with ratio performance guarantee In this section we present a heuristic algorithm for the job shop problem with a single bottleneck machine. We start with considering the ow shop problem; the generalization of our approach to the job shop is quite straightforward. Most of heuristic algorithms for the F m 1 jm j 6 2jC max problem use a certain version of the arti cial F 2jjC max problem and apply the Johnson rule [10] to determine the sequences of jobs on all machines, including the bottleneck machine, (see e.g., [6]). Our algorithm for the ow shop problem starts with an arbitrary schedule and transforms it into another schedule with the makespan that is at most 3/2 times the optimal value. Let S be an optimal schedule for the J m 1 jm j 6 2jC max problem. The so-called machine-based lower bound on the value of the makespan is apparent: C max S P maxfa N 1 [ H 1 ; a N 2 [ H 2 ;...; a N m [ H m ; b N [ H g: 2 For the F m 1 jm j 6 2jC max problem for processing the jobs of set N; let an optimal schedule be denoted by SN. In this case (2) becomes C max S N P max f a N 1 ; a N 2 ;...; a N m ; b N g: 3 For a job u 2 N and for any set U fj j a j P a u ; j 2 Ng the lower bound C max S N P a u b U 4 holds, since in any feasible schedule no job of set U can be started on B earlier than time a u. We now present an algorithm that creates a ow shop schedule with the makespan that is at most 3/2 times the optimal value. Algorithm RG. Input: An instance of the F m 1 jm j 6 2jC max problem with the set N of jobs. Output: Schedule S N. 1. For each k; 1 6 k 6 m; take an arbitrary permutation p k N k of the jobs of set N k. Run Algorithm FSN taking these permutations as the input. Call the found schedule S 0 : 2. In schedule S 0 ; nd a critical job w and de ne s as the starting time of job w on machine B. De ne the index set M 0 such that M 0 ˆ fk j 1 6 k 6 m; a N k P sg: 3. For each k 2 M 0 do the following. Find the job w k 2 N k for which R wk ;A S 0 < s and C wk ;A S 0

7 I.G. Drobouchevitch, V.A. Strusevich / European Journal of Operational Research 123 (2000) 229± P s: Find the set Nk 0 of jobs, which precede job w k in the sequence p k N k : De ne Nk 00 as the set of jobs that follow job w k in the sequence p k N k. 4. For each k; 1 6 k 6 m; de ne the sequence r k N k ˆ p k N 00 k ; p k N 0 k ; w k if k 2 M 0 ; otherwise de ne r k N k ˆp k N k. Run Algorithm FSN taking these permutations as the input. Call the found schedule S 1 : 5. For each k; 1 6 k 6 m; de ne the sequence v k N k ˆ w k ; p k N 0 k ; p k N 00 k if k 2 M 0 ; otherwise de ne v k N k ˆp k N k. Run Algorithm FSN taking these permutations as the input. Call the found schedule S 2 : 6. Among the found schedules S 0 ; S 1 and S 2 ; determine the one with the smallest value of the makespan. Call this schedule S N. Output S N and stop. The running time of Algorithm RG does not exceed an O nm. The theorem below analyses its worst-case performance. Theorem 1. Algorithm RG generates a schedule S N for the F m 1 jm j 6 2jC max problem such that the bound C max S N C max S N holds and this bound can not be reduced. 5 Proof. We start with considering schedule S 0 associated with an arbitrary collection of sequences of the rst-stage operations. Schedule S 0 is terminated on the bottleneck machine B. We assume that the other machines are indexed in such a way that a N 1 P a N 2 P P a N m : 6 Let w be the critical job in S 0 ; and the processing of this job on machine B starts at time R w;b S 0 ˆs. We identify the largest index q such that a N q P s; in other words M 0 ˆf1; 2;...; qg (see step 2). Let the jobs w k and the sets Nk 0 00 and Nk ; k 2 M 0 ; be de ned as in step 3. De ne the sets W ˆfw 1 ; w 2 ;...; w q g; N 00 ˆ [q N 0 ˆ N n N 00 [ W : N 00 k ; kˆ1 Notice that in schedule S 0 the sequence of jobs on B is found by Algorithm FSN. Because of the greedy nature of that algorithm, any job that is completed on a rst-stage machine by time s is also completed on B by time s; otherwise, job w would not be the critical job. This implies that machine B after time s processes only the jobs of sets W and N 00. By de nition of job w k ; we have that a Nk 0 < s and a Nk 0 a w k P s, which imply that C max S 0 6 a N 0 k a w k b W bn 00 7 is valid for each k 2 M 0. Notice that for the critical job w k ˆ w the relation above turns into the equality. If a N 0 k a w k C max S N for some k 2 M 0, then (7) gives (5) due to (3). Therefore, in the rest of this proof we assume that a N 0 k a w k > 1 2 C max S N ; k 2 M 0 : These inequalities and (3) imply that an 00 1 k < 2 C max S N ; k 2 M 0 : 8 Consider now schedule S 1 found in step 4, that is associated with the collection of sequences r 1 N 1 ; r 2 N 2 ;...; r m N m. Let x denote the critical job in schedule S 1. For schedule S 1, suppose that for some r 2 M 0 there exists a job u 2 Nr 00 that is started on B no earlier than the critical job x. Recall that the index set M 0 and the set Nr 00 of jobs are de ned with respect to schedule S 0. Then C max S 1 6 R x;b S 1 b N 6 an 00 r b N ; which yields (5) due to (3) and (8). Therefore, we assume that all jobs that start on B no earlier than the critical job belong to the set W [ N 0. We show that the inequality C max S 1 6 a N 1 b W 9 holds. For schedule S 1 ; determine the set Q consisting of all jobs in N 0 that start on B no earlier

8 236 I.G. Drobouchevitch, V.A. Strusevich / European Journal of Operational Research 123 (2000) 229±240 than the critical job. If Q ˆ;, then (9) obviously holds. Thus, we further assume that Q 6ˆ ;. In schedule S 0 all jobs of set Q are completed on B before time s. Find the job y, which is the rst job of set Q to be processed on B in schedule S 0.It follows that C y;a S 0 b Q 6 s: On the other hand, in schedule S 1 job x is critical and y starts on B no earlier than x. This implies that C y;a S 1 P C x;a S 1, so that C max S 1 6 C y;a S 1 b Q b W : If job y belongs to set Nr 0 for some r 2 M 0, then we obtain that C y;a S 1 ˆaNr 00 Cy;A S 0 ; otherwise, C y;a S 1 ˆC y;a S 0. In the former case, we have that C max S 1 6 an 00 r s b W 6 a Nr b W ; while in the latter case, we have that C max S 1 6 s b W 6 a N 1 b W : This proves (9). Now, if b W 6 1 C 2 max SN, then (5) obviously holds. Hence, in the remainder of this proof we assume that b W > 1 2 C max S N : 10 Consider now schedule S 2 found in step 5, that is associated with the collection of sequences v 1 N 1 ; v 2 N 2 ;...; v m N m. Let z denote the critical job in schedule S 2. For schedule S 2 ; de ne W 0 as the set consisting of all jobs of set W that are completed on B no earlier than the critical job z. IfW 0 ˆ;, then C max S 2 6 C z;a S 2 b N n W 6 a N 1 b N n W ; so that (5) holds due to (3) and (10). Otherwise, let w p be a job such that a wp ˆ min a wj j w j 2 W 0 : Each job in set W 0 starts on B no earlier than time C wp ;A S 2 ˆa wp : Since z is critical in S 2,we have that a wp P C z;a S 2, so that C max S 2 6 a wp b W 0 b N n W : Applying (4) with u ˆ w p and U ˆ W 0 yields a wp b W 0 6 C max SN. Therefore, C max S C 2 max SN due to (10). This proves the desired bound (5). To see that 3/2 is the best possible ratio for Algorithm RG, consider the following instance of the F m 1 jm j 6 2jC max problem. Let Z be a su ciently large positive number, e.g., Z m. The set of jobs N 1 contains two jobs 1 and 2 with a 1 ˆ Z; b 1 ˆ Z m 1; a 2 ˆ Z; b 2 ˆ 0: The set N m consists of Z m 4 jobs numbered by the integers 3m 3, 3m 2,...; Z 2m such that a 3m 3 ˆ Z; b 3m 3 ˆ 1; and a j ˆ b j ˆ 1; j ˆ 3m 2;...; Z 2m 1; a Z 2m ˆ m 2; b Z 2m ˆ 0: If m > 2, then each set N k ; 2 6 k 6 m 1; consists of three jobs 3k 3; 3k 2 and 3k 1 such that a 3k 3 ˆ Z; b 3k 3 ˆ 1; a 3k 2 ˆ b 3k 2 ˆ 1; a 3k 1 ˆ Z 1; b 3k 1 ˆ 0: Notice that zero processing times are understood as very small positive values which can be neglected. Algorithm RG may start with schedule S 0 associated with the permutations p 1 N 1 ˆ 1; 2 ; p k N k ˆ 3k 3; 3k 2; 3k 1 ; k ˆ 2;...; m 1; p m N m ˆ 3m 3; 3m 2;...; Z 2m : In S 0, for each k; 1 6 k 6 m; a job, which is sequenced to be rst on machine A k ; is completed on that machine at time Z. Then machine B is idle till time Z, and processes the jobs 1; 3; 7;...; 3m 3 in any order in the time interval Z; 2ZŠ. By time 2Z, every machine A k completes the processing of the remaining jobs. These jobs can be processed on B without idle time, starting at time 2Z. Thus, C max S 0 ˆ3Z. Moreover, s ˆ Z, and w 1 ˆ 1; w k ˆ 3k 3; k ˆ 2;...; m. The structure of schedule S 0 implies that Nk 0 ˆ; for each k; 1 6 k 6 m; and therefore, schedule S 2 will coincide with S 0. Schedule S 1 is associated with the permutations r 1 N 1 ˆ 2; 1 ; r k N k ˆ 3k 2; 3k 1; 3k 3 ; k ˆ 2;...m 1; r m N m ˆ 3m 2;...; Z 2m; 3m 3 : In this schedule, for

9 I.G. Drobouchevitch, V.A. Strusevich / European Journal of Operational Research 123 (2000) 229± each k; 2 6 k 6 m; a job, which is sequenced to be rst on machine A k ; is completed on that machine at time 1. Machine B processes the jobs 4; 7;...; 3m 2 in any order in the time interval 1; mš. Machine A m processes Z m 1 jobs numbered by 3m 1; 3m;...; Z 2m 1 in the time interval 1; Z m 2Š. These jobs are processed in the same order on B in the interval m; Z 1Š. Job 2 on machine A 1 is completed at time Z. Every machine A k ; 2 6 k 6 m 1, in the interval 1; ZŠprocesses job 3k 1, while job Z 2m completes on machine A m at time Z. All these jobs for which the rst-stage operation is completed at time Z, have a dummy second-stage operation, supplying no load for machine B: After time Z every machine A k processes a job with the processing time Z, so that all these jobs are completed simultaneously at time 2Z. These jobs can be processed on B without idle time, starting at time 2Z. Thus, C max S 1 ˆ3Z. On the other hand, there exists an optimal schedule SN with C max SN ˆ2Z 1. This schedule is associated, e.g., with the collection of permutations u 1 N 1 ˆ 1; 2 ; u k N k ˆ 3k 2; 3k 3; 3k 1 ; k ˆ 2;...; m 1; u m N m ˆ 3m 2;...; Z 2m 1; 3m 3; Z 2m. The jobs with dummy second-stage operations are sequenced last on the corresponding machine. It can be veri ed that machine B starts at time 1 and has no idle time. Thus, we have C max S N C max SN ˆ 3Z 2Z 1 ; and this ratio approaches 3/2 as Z grows. This proves the theorem. Observe that in fact Algorithm RG takes an arbitrary schedule as schedule S 0 and outputs the best of three schedules. Intuitively, the performance of the algorithm could be improved if a speci c schedule S 0 was chosen as a starting point. So far we have failed to nd a suitable rule for choosing S 0, or rather for de ning the initial permutations p k N k ; k ˆ 1; 2;...; m. The proof of the tightness of the bound (5 ) in particular implies that a ratio of 3/2 will not be reduced if each sequence p k N k is de ned as the Johnson permutation of the jobs of set N k : We now turn back to the original J m 1 jm j 6 2jC max problem. Algorithm RG can be easily modi ed for nding a heuristic schedule S H for the ow shop problem of processing the jobs of set H, for which the rst-stage operations are processed on the bottleneck machine. Thus, by running Algorithm RG and then its modi cation, we may nd two schedules S N and S H : Notice that the makespan of each of these schedules is at most than 3/2 times larger than the optimal value for the corresponding ow shop problem, and, therefore, than the optimal job shop makespan. These schedules can be then combined as described in Algorithm JS. If in the resulting schedule the makespan is determined by the workload of some machine, then the schedule is optimal due to (2). Otherwise, the overall makespan is equal to the makespan of one of the heuristic ow shop schedules S N or S H. This implies the following statement. Theorem 2. For the J m 1 jm j 6 2jC max problem, a schedule S such that C max S =C max S 6 3=2 can be found in no more than O nm hm time. The ratio bound of 3/2 is tight. 6. Heuristic with absolute performance guarantee In this section we study heuristics that provide absolute worst-case performance guarantees. The algorithm given below is an extension of the heuristic described in [13]. The latter algorithm is designed for the F m 1 jm j 6 2jC max problem with m ˆ 2. It creates a ow shop schedule in which the jobs are sorted in non-increasing order of the ratio b j =a j. We demonstrate that a similar approach is applicable to a more general situation. Algorithm AG. Input: An instance of the J m 1 jm j 6 2jC max problem. Output: A heuristic schedule S U. 1. For each job j 2 N [ H; de ne c j ˆ bj=a j if a j > 0; Z otherwise; where Z > maxfb j =a j j a j > 0; j 2 N [ Hg.

10 238 I.G. Drobouchevitch, V.A. Strusevich / European Journal of Operational Research 123 (2000) 229± For each k; 1 6 k 6 m, nd the sequence u k N k by sorting the jobs of set N k in non-increasing order of c j. Find the sequence w k H k by sorting the jobs of set H k in non-decreasing order of c j. 3. Run Algorithm JS and call the found schedule S U. Stop. The running time of Algorithm AG does not exceed O n log n h log h n h m. The main advantage of the ordering of the jobs as in step 2 is that useful relations between the ratios of certain partial sums can be derived. These relations are established and explored below. Lemma 1. Let u k N k be a sequence found in step 2 of Algorithm AG. Given a job u 2 N k, define the set N 1 k of all jobs that precede u in u k N k : If a N 1 k > 0, then b Nk 1 a Nk 1 P b N k 2 a Nk 2 ; where N 2 k ˆ N k n N 1 k : 11 Proof. Without loss of generality, assume that the jobs in N k are numbered in such a way that u k N k ˆ 1; 2;...; r. Therefore, Nk 1 ˆf1; 2;...; u 1g and Nk 2 ˆfu; u 1;...; rg. Let s be the rst job in u k N k with a s > 0: The inequality a Nk 1 > 0 implies that s exists and belongs to N k 1. It is obvious that P s jˆ1 b j P s jˆ1 a P b s : 12 j a s It is well known that for any four numbers c P 0; d > 0; e P 0; f > 0 the inequality c=d P e=f implies c d P c e d f P e f : Since in u k N k the jobs s; s 1;...; r are sorted in non-increasing order of the ratios b j =a j, it follows that b s a s P b s b s 1 a s a s 1 P P b s b s 1 b u 1 a s a s 1 a u 1 P b u 1 a u 1 : On the other hand, b u P b u b u 1 P... a u a u a u 1 P b u b u 1... b r a u a u 1... a r : Combining (12)±(14), we obtain the desired inequality (11). This proves the lemma. Recall that P denotes the largest of the machine workloads, while p max stands for the largest processing time of an operation. We estimate the makespan of the found schedule in terms of P and p max. Theorem 3. For the J m 1 jm j 6 2jC max problem Algorithm AG generates schedule S U such that C max S U 6 P p max : 15 Proof. Without loss of generality, we assume that C max S U is determined by the length of the ow shop schedule for processing the jobs of set N. Similar to the proof of Theorem 1, let the machines be numbered so that (6) holds. If for some k; 1 6 k 6 m; the inequality C max S U 6 a N k p max holds, then the bound (15) follows immediately. Thus, in the remainder of this proof we assume that C max S U > a N k p max ; k ˆ 1; 2;...; m: 16 To prove the theorem we need to show that C max S U 6 b N [ H p max : 17 Let w be the critical job in S U ; and the processing of this job on machine B starts at time R w;b ˆ s. We identify the largest index q such that a N q P s: Similarly to step 3 of Algorithm RG, for each k, 1 6 k 6 q; we nd the job w k 2 N k that starts on A k before time s and nishes on that machine no earlier than time s, and de ne Nk 1 to be the subset of N k consisting of the jobs that precede job w k in

11 I.G. Drobouchevitch, V.A. Strusevich / European Journal of Operational Research 123 (2000) 229± the permutation u k N k ; also de ne Nk 2 ˆ N k n Nk 1. In schedule S U after time s machine B processes only the jobs of the sets Nk 2 ; k ˆ 1; 2;...; q; with no intermediate idle time, i.e., C max S U ˆs Xq kˆ1 b N 2 k : 18 The inequality s 6 p max leads to the desired relation (17). Therefore, we assume that s > p max : X q kˆ1 To prove (17), it su ces to show that b N 1 k P s p max: By de nition of job w k ; we have a N 1 k a w k P s, which implies that s p max 6 a N 1 k ; k ˆ 1; 2;...; q: 21 Due to (19), the inequality a Nk 1 > 0 holds for each k; 1 6 k 6 q. We now apply Lemma 1 with u ˆ w k to each set N k ˆ Nk 1 [ N k 2 ; 1 6 k 6 q, to obtain b N 1 k P b N 2 k a N 1 k a N 2 k : We use a Nk 2 ˆa N k a Nk 1 and (21) to derive b N 1 k P s p max b Nk 2 ; k ˆ 1; 2;...; q: a N k s p max Summing up and using (16) we get X q kˆ1 b N 1 k P s p max Xq b Nk 2 a N kˆ1 k s p max > s p max P q kˆ1 b N k 2 : C max S U s The required inequality (20) directly follows from (18). This proves the theorem. Notice that the bound (15) cannot be improved in the following sense: for any a < 1 there exists an instance of our problem such that C max S > P ap max : This can be seen from considering the trivial instance of the job shop problem, in which H ˆ;and N k ˆfkg; a k ˆ b k ˆ 1; k ˆ 1; 2;...; m: Indeed, here P ˆ m, p max ˆ 1 and C max S ˆ m Conclusion In this paper we have described and analyzed heuristic algorithms for the two-stage job shop scheduling problem with a single bottleneck machine. The obtained worst-case analysis results improve those known for less general scheduling models. One of possible directions of future research is to extend the obtained results to the general twostage job shop. We suspect that the non-approximability result by Williamson et al. [19] does not hold in this case, so that fast heuristics with a worst-case ratio of 5/4 or better are likely to exist. It is also worth studying how the techniques considered in this paper can be applied to the related problems, such as the ow shop with parallel machines or the assembly problem. Acknowledgements This research was partly supported by the International Association for the Promotion of Cooperation with Scientists from the Independent States of the Former Soviet Union, INTAS Ext. We are grateful to the anonymous referees whose comments have contributed to improving the presentation. References [1] B. Chen, Analysis of classes of heuristics for scheduling a two-stage ow shop with parallel machines at one stage, Journal of the Operational Research Society 46 (1995) 234±244. [2] B. Chen, C.A. Glass, C.N. Potts, V.A. Strusevich, A new heuristic three-machine ow shop scheduling, Operations Research 44 (1996) 891±898. [3] I.G. Drobouchevitch, V.A. Strusevich, Heuristics for short route job shop scheduling problems, Mathematical Methods of Operations Research 48 (1998) 359±375.

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